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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.2 Integration by Parts

Focus on choose u and dv, logarithmic integrands, repeated integration by parts.

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Ideas in this lesson

01Choose u and dvWorked steps · five questions02Logarithmic integrandsWorked steps · five questions03Repeated integration by partsWorked steps · five questions
Idea 1 of 3

Choose u and dv

Integration by parts comes from the product rule. Choose uu so differentiating it simplifies the remaining integral.

Worked example 1

Evaluate ∫012xex dx\int_{0}^{1} 2xe^{x}\,dx.
  1. u=x,du=dxu = x, du = dx; dv=2exdx,v=2exdv = 2e^{x}dx, v = 2e^{x}
  2. ∫2xexdx=2xex−2ex+C\int 2xe^{x} dx = 2xe^{x} - 2e^{x} + C
  3. [2ex(x−1)]∣01=0−(−2)=2[2e^{x}(x - 1)]\big|_{0}^{1} = 0 - (-2) = 2

Answer: 22

Worked example 2

Evaluate ∫0π3xsin⁡x dx\int_{0}^{\pi } 3x \sin x\,dx.
  1. u=x,du=dxu = x, du = dx; dv=3sin⁡xdx,v=−3cos⁡xdv = 3 \sin x dx, v = -3 \cos x
  2. ∫3xsin⁡xdx=−3xcos⁡x+3sin⁡x+C\int 3x \sin x dx = -3x \cos x + 3 \sin x + C
  3. F(π)−F(0)=3π−0=3π≈9.42F(\pi ) - F(0) = 3\pi - 0 = 3\pi \approx 9.42

Answer: 3π≈9.423\pi \approx 9.42

Idea 2 of 3

Logarithmic integrands

For ln⁡x\ln x, use dv=dxdv = dx and u=ln⁡xu = \ln x. The derivative becomes 1x\frac{1}{x}.

Worked example 1

Evaluate ∫1e2ln⁡x dx\int_{1}^{e} 2 \ln x\,dx.
  1. u=ln⁡x,du=1xdxu = \ln x, du = \dfrac{1}{x}dx; dv=2dx,v=2xdv = 2dx, v = 2x
  2. F(x)=2(xln⁡x−x)F(x) = 2(x \ln x - x)
  3. F(e)−F(1)=2[e(0)+1]=2F(e) - F(1) = 2[e(0) + 1] = 2

Answer: 22

Worked example 2

Evaluate ∫1e23ln⁡x dx\int_{1}^{e^{2}} 3 \ln x\,dx.
  1. u=ln⁡x,du=1xdxu = \ln x, du = \dfrac{1}{x}dx; dv=3dx,v=3xdv = 3dx, v = 3x
  2. F(x)=3(xln⁡x−x)F(x) = 3(x \ln x - x)
  3. F(e2)−F(1)=3[e2(1)+1]=3(e2+1)≈25.17F(e^{2}) - F(1) = 3[e^{2}(1) + 1] = 3(e^{2} + 1) \approx 25.17

Answer: 3(e2+1)≈25.173(e^{2} + 1) \approx 25.17

Idea 3 of 3

Repeated integration by parts

A polynomial times an exponential may require repeated applications until the polynomial differentiates away.

Worked example 1

Evaluate ∫012x2ex dx\int_{0}^{1} 2x^{2}e^{x}\,dx.
  1. u=x2,du=2xdxu = x^{2}, du = 2x dx; dv=2exdx,v=2exdv = 2e^{x}dx, v = 2e^{x}
  2. ∫2x2exdx=2x2ex−4∫xexdx\int 2x^{2}e^{x}dx = 2x^{2}e^{x} - 4\int xe^{x}dx
  3. F(x)=2ex(x2−2x+2)F(x) = 2e^{x}(x^{2} - 2x + 2)
  4. F(1)−F(0)=2e−4=2(e−2)≈1.44F(1) - F(0) = 2e - 4 = 2(e - 2) \approx 1.44

Answer: 2(e−2)≈1.442(e - 2) \approx 1.44

Worked example 2

Evaluate ∫013x2ex dx\int_{0}^{1} 3x^{2}e^{x}\,dx.
  1. u=x2,du=2xdxu = x^{2}, du = 2x dx; dv=3exdx,v=3exdv = 3e^{x}dx, v = 3e^{x}
  2. ∫3x2exdx=3x2ex−6∫xexdx\int 3x^{2}e^{x}dx = 3x^{2}e^{x} - 6\int xe^{x}dx
  3. F(x)=3ex(x2−2x+2)F(x) = 3e^{x}(x^{2} - 2x + 2)
  4. F(1)−F(0)=3e−6=3(e−2)≈2.15F(1) - F(0) = 3e - 6 = 3(e - 2) \approx 2.15

Answer: 3(e−2)≈2.153(e - 2) \approx 2.15

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.2,p8.2, p. 515515. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.